Biology2 - Esther Cha Chemistry 162 Siegel Rutchem.rutgers.edu/sakai Mastering chem CHEM162508BeckDay Coordinator [email protected] Dr

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Esther Cha Chemistry 162 Siegel 01-23-08 Rutchem.rutgers.edu/sakai Mastering chem- CHEM162508BeckDay Coordinator: [email protected] Dr. Donald Siegel Wright labs 135 (732-445-5205) Clicker Setup: On/* until “setup menu”/triangle until “ID”/ type in RUID + Chapter 12 12.1: Types of Solutions Solvent: determines state of matter of solution; usually greatest amount Solute: substances dissolved in solvent Proportion of solute dilute solution v. concentrated solution Solutions are not limited to the liquid phase Physical Properties of Solutions- Solutions are homogenous (same throughout) Air is a solution: N is the solvent, and the other gases are solutes 12.2: Solution concentration Measure concentration by: 1) % concentration (w/w: weight/weight- usually used; v/v: volume/volume; w/v: weight/volume) v/v: used for liquids w/v: used for dilute solutions 2) Ppm- parts per million gaseous solutions: based on numbers of molecules or on volumes liquid solutions: based on masses 3) X: mole fraction (moles solute/moles solution) x of a component = moles of component/total moles of solution components has no unit; is dimensionless sum of mole fractions of all components is 1 4) M: molarity varies with temperature
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Esther Cha Chemistry 162 Siegel 5) M: molality (moles of solute/kg of solvent) concentration based on mass only, not volume concentration is independent of temperature, like percent by mass (w/w) e.g.: Find the molality of a 1.00% w/w solution of CH3CH2OH (ethanol) 1.00g CH3CH2OH x 1 mol/44.07g = 2.12x 10^(-2) mol 2.17 x 10^(-2) mol/0.10000 kg = 0.217 m *Really 0.001 solute, 0.099 solvent (but insignificant detail) Relationship between M & m - For dilute solutions, M ≈ m e.g.: 3.75 M H2SO4 has density 1.230 g/mL. What is the molality? 1 L weighs 1230 g 1 L has 3.75 mol H2SO4 = 368g 1230-368 = 862 g H20 m = 3.75 mol/0.862 kg = 4.35 m 12.3 Energetics of Solution Formation Enthalpy of Solution- (energetics of solutions) 3 Hypothetical steps to solution formation: 1) Enthalpy to overcome intermolecular forces in solvents (endothermic) = separation of solvent molecules, ΔH1 2) Enthalpy to overcome intermolecular forces in solutes (endothermic, energy expended) = separation of solute molecules, ΔH2 Magnitude depends on compound Large for polar, ionic compounds 3) Enthalpy of solvation (“hydration” when water is the solvent) (exothermic, energy released) = all separated molecules mix randomly and become a solution, ΔH3
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This note was uploaded on 04/03/2008 for the course CHEM 162 taught by Professor Siegal during the Spring '08 term at Rutgers.

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Biology2 - Esther Cha Chemistry 162 Siegel Rutchem.rutgers.edu/sakai Mastering chem CHEM162508BeckDay Coordinator [email protected] Dr

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